I follow my professors derivation. However, she expands the term: ##\vec { \nabla } \times \left( \vec { m } \times \vec { r } \right) =2 \vec m## by just using the determinate. I figured I would go with the less messy "BAC CAB" like she uses on another term. However, I can't get ##2 \vec m## by using BAC CAB. I get ##3 \vec m## instead. I have tried both ways using spherical and cartesian. Does not ##\nabla \cdot \vec m=0##? If I can show that ##\vec { r } \left(\vec \nabla \cdot \vec m \right)=-\vec m##. I would be home free, but I think it should equal zero.